Lecture452 18 10.Pdf

University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2010 Lecture 18 8/02/10 A. Review of the Gibbs-Duhem Equation • The Gibbs-Duhem equation has the form (neglecting surface tension effects): SdT− VdP+=∑ njj dµ 0 j • At constant T and P and for a binary solution the Gibbs-Duhem equation has the form n1 nd11µ +=⇒=− nd 2µµµ 20 d 2 d 1 n2 • For practical calculations the Gibbs-Duhem eqn is commonly expressed in terms of activities and molality is used instead of mole fraction. The G-D eqn now has m1 the form: daln21=− da ln m2 • In principle, one can obtain the activity a2 of the solution by integrating both side m1 of the Gibbs-Duhem equation daln21=− da ln . In practice such integrals m2 have problems in the dilute limit where m2 approaches zero. At this point the integrant blows up and accurate results are difficult to obtain. • A more stable integration is obtained using the molal osmotic coefficient: m1 φ =− ln a1 m2 md222lnγ = mdφφ+− dm 22 dm φ −1 ∴=+ddlnγφ22 dm m2 • This equation is integrated from m2=0 where φ=1 and γ2=0, to m2, giving the activity coefficient for the solute at m2: m2 φ −1 lnγφ=+−dm′ 1 22∫ 0 m2′ • Consider a binary solution composed of a nonvolatile solute and a volatile solvent in equilibrium with its (ideal) vapor. We have PP11⎛⎞ 00=⇒aa11ln = ln ⎜⎟ PP11⎝⎠ • We then define the molal osmotic coefficient in terms of the vapor pressure of the solvent : mmP111⎛⎞ φ =−lna1 =− ln ⎜⎟0 mmP221⎝⎠ and we evaluate the activity of a non-volatile solute at some molality m2 from the vapor pressure data obtained between m2=0 and m2. B. Colligative Properties: Osmotic Pressure • A colligative property is a property of a solution that depends on a given amount of solvent, but not on the kind of solute present. • Measurements of colligative properties like vapor pressure, freezing point depression, and osmotic pressure, are presented in the text as probing solute concentration. But as the vapor pressure example above illustrates, measuring the vapor pressure of a volatile solvent enables measurement of the activity coefficient of a non-volatile solute. • Osmotic Pressure: Osmotic pressure is an example of a colligative property. Osmotic pressure is typically described for ideal solutions where it can be directly related to solute concentration. Osmotic pressure is the best way to obtain activity coefficients for polymeric solute. • We consider the diagram below. A compartment with two stacks is separated by a semipermeable membrane. On one side of the membrane is a solvent 1 (e.g. water). On the other side is a solution consisiting of the same solvent 1 and a solute 2. The membrane is permeable to the solvent but no to the solute. • To reach equilibrium the free energy of the solvent must be equal on both sides of the membrane i.e. µµ11()left= ( right ) 0 • In the left-hand-compartment is pure solvent so µ11(left) = µ . 0 • In the right-hand-compartment is a solution so µµ111(right) =+ RTln a • The only way that this system can reach equilibrium is for solvent to flow from the left-hand compartment through the membrane to the right-hand compartment. However, in doing so the volume of the solution increase and a column of water rise in the right-hand stack. The difference between the liquid heights in the right-hand and left hand stacks is h. • To push water in the right-hand stack a pressure π must be exerted in excess of the atmospheric pressure. To push a column of solution with density ρ up a height h requires π = ρgh • To relate the pressure p to properties of the solution consider that to produce this pressure the free energy of the solution must increase by 1+π ∆=µµd = VdP − SdT = V dP(0) if dT = ∫∫() ∫1 1 00 • The equilibrium condition is µ11(left) = µµµµ( right) or 111=+ RTln a +∆ 1+π =+µ 00RTaln + VdP =+µπ RTaV ln + 111111∫ 1 • Rearranging πV11= − RTln a≈ − RT lnχ 1= − RT lnb1− χ 2g n2 n2 • If the solution is dilute thenχ 2 = ≈<<1 nn12+ n1 • lnb 1−≈−xxifxg << 1 n2 • Therefore πχππVRT12=−lnb1 −g ≈ RT ⇒Vn11 = V = n 2 RT n1 n • Rearrange to ππVnRT=⇒==2 RT c RT 2 V 2 • Van’t Hoff’s Law: In the limit of infinite dilution the osmotic pressure π is defined as π = c2 RT . Example: Calculate the osmotic pressure associated with a 0.1M sucrose solution at T=298K. Solution: π ==c2 RTbgb01.. M 0 0821 L⋅ atm / moles⋅ K gbg 298 K= 2 .. 44 atm Note: 1 atm of pressure would push a column of mercury 0.76 m high. Therefore the osmotic pressure described above would support a column of mercury 1.85 m high. The same pressure would support a column of water 25 meters high (i.e. 78 feet). Comment: The concept of osmotic pressure is frequently misunderstood. When we say a solution has an osmotic pressure of 2.44 atm we do not imply that the pressure in the solution is 2.44 atm. By osmotic pressure we mean the extra pressure that must be applied to the solution to bring it into equilibrium with pure solvent across a semipermeable membrane. An illustration is shown below. At the left, below is a double compartmented volume, separated by a semipermeable membrane. If each side of the membrane is occupied by pure water, the pressures exerted on each compartment must be equal (Pleft=Pright) in order for the water transfers between the compartments to be equal, that is, for the two sides to be in equilibrium across the membrane. If the compartment on the right is filled with solution, an additional pressure, the osmotic pressure OP, must be exerted on the right-hand compartment to bring it into equilibrium with the solvent in the let-hand compartment. Another example: Suppose a double-stacked container is divided into two compartments by a membrane (see diagram on page 1 of this lecture). The left-hand compartment is filled with water and the right hand compartment is filled with a solution composed of 20 grams of hemoglobin in 1 liter of water. At equilibrium, the liquid level in the right-hand stack (i.e. the solution side) is 77.8 mm above the level in the left-hand stack. The temperature is 298K. What is the osmotic pressure of the hempglobin solution and what is the molecular weight of hemoglobin? Assume the solution density is equal to the density of pure water. Solution: πρ==ghc1033 kg/./. mhc 9 8 m s 2hb 0 0778 mg = 762 ./ 46 N m 2 Using the van’t Hoff Law… 2 002. kg F 1 I π ===762./ 46Nm cRT F I 8./ 31J mole⋅ K 298 K 2 G 3 JG Jb gb g H 0. 001m KH M 2 K Rearrange: chc20kg/. m32211 8 31 kg⋅⋅⋅ m s−− K ⋅ mole − hb 298 Kg M = = 65040./kg mole 2 762. 46kg⋅⋅ m−−12 s.

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